如何在Python通讯录代码中实现联系人的精准删除？

问题根源 #

原代码的删除逻辑使用if name not in line:判断，只要行文本中包含输入的姓名片段就会被过滤删除。比如当通讯录里有"J"和"James"两个联系人时，删除"J"会同时删掉包含"J"的"James"条目，无法实现精准匹配。

修复思路 #

由于联系人条目格式固定为姓名 : 电话号码，我们可以通过拆分每行内容提取出完整的联系人姓名，再与要删除的姓名做精确相等比对，只过滤掉姓名完全匹配的条目。

关键代码修改 #

修正后的删除模块

elif q == '4':
    target_name = input('Enter name to delete from contact book: ')
    print('\n')
    with open("contact.txt", "r+") as f:
        lines = f.readlines()
        f.seek(0)
        for line in lines:
            # 去除换行符与前后空白，跳过空行
            cleaned_line = line.strip()
            if not cleaned_line:
                continue
            # 按固定分隔符拆分，只拆分一次避免电话号码含冒号出错
            contact_name, _ = cleaned_line.split(' : ', 1)
            # 仅当姓名完全不匹配时，保留该行
            if contact_name != target_name:
                f.write(cleaned_line + '\n')
        f.truncate()

额外优化：修复搜索模块的逻辑漏洞

原搜索模块会在第一行不匹配时直接终止循环，漏掉后续可能的匹配项，同步修复如下：

elif q == '2':
    with open('contact.txt', 'r') as f:
        search_key = input('Search: ')
        found = False
        for line in f:
            cleaned_line = line.strip()
            if not cleaned_line:
                continue
            contact_name, phone = cleaned_line.split(' : ', 1)
            # 支持姓名或电话模糊搜索
            if search_key in contact_name or search_key in phone:
                print(f"{contact_name} : {phone}\n")
                found = True
        if not found:
            print('Contact not found \n')

完整修正代码 #

while True:
    q = input('Press 1 to add, 2 to search, 3 to display, 4 to delete, 5 to clear contact book and 6 to quit: ')
    if q == '1':
        with open('contact.txt', 'a') as f:
            name = input('Name: ')
            phone= input('Phone: ')
            f.writelines((name, ' : ', phone, '\n'))
            print('\n')

    elif q == '2':
        with open('contact.txt', 'r') as f:
            search_key = input('Search: ')
            found = False
            for line in f:
                cleaned_line = line.strip()
                if not cleaned_line:
                    continue
                contact_name, phone = cleaned_line.split(' : ', 1)
                if search_key in contact_name or search_key in phone:
                    print(f"{contact_name} : {phone}\n")
                    found = True
            if not found:
                print('Contact not found \n')

    elif q == '3':
        with open('contact.txt', 'r') as f:
            print('The current entries in the contact book are \n')
            for line in f:
                print(line, end='')
            print('\n')

    elif q == '4':
        target_name = input('Enter name to delete from contact book: ')
        print('\n')
        with open("contact.txt", "r+") as f:
            lines = f.readlines()
            f.seek(0)
            for line in lines:
                cleaned_line = line.strip()
                if not cleaned_line:
                    continue
                contact_name, _ = cleaned_line.split(' : ', 1)
                if contact_name != target_name:
                    f.write(cleaned_line + '\n')
            f.truncate()

    elif q == '5':
        with open('contact.txt', 'w+') as f:
            f.truncate()
                
    else:
        break

内容的提问来源于stack exchange，提问作者zeitgeist_7